The integral $\int \frac { 1 } { x ^ { 2 } + 2 x + 10 } d x$ is

A) $\frac { \tan ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) } { 9 } + C$
B) $\frac { \tan ^ { - 1 } \left( \frac { x - 1 } { 3 } \right) } { 9 } + C$
C) $- \frac { \tan ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) } { 3 } + C$
D) $\frac { \tan ^ { - 1 } \left( \frac { x - 1 } { 3 } \right) } { 3 } + C$
E) $\frac { \tan ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) } { 3 } + C$

Question

The integral $\int \frac { 1 } { x ^ { 2 } + 2 x + 10 } d x$ is

A) $\frac { \tan ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) } { 9 } + C$
B) $\frac { \tan ^ { - 1 } \left( \frac { x - 1 } { 3 } \right) } { 9 } + C$
C) $- \frac { \tan ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) } { 3 } + C$
D) $\frac { \tan ^ { - 1 } \left( \frac { x - 1 } { 3 } \right) } { 3 } + C$
E) $\frac { \tan ^ { - 1 } \left( \frac { x + 1 } { 3 } \right) } { 3 } + C$

Quizplus · Accepted Answer

The integral can be solved by completing the square in the denominator to get $x^2 + 2x + 10 = (x+1)^2 + 9$, and then using the formula for the integral of $1/(a^2 + x^2)$, which results in $\frac{1}{a} \tan^{-1}(x/a) + C$. Here, $a = 3$, leading to the correct answer $\frac{\tan^{-1}((x+1)/3)}{3} + C$.

The Integral ∫1x2+2x+10dx\int \frac { 1 } { x ^ { 2 } + 2 x + 10 } d x∫x2+2x+101​dx

The Integral $\int \frac { 1 } { x ^ { 2 } + 2 x + 10 } d x$